Respuesta :
Solution :
Given : Mass of ladder = 10 kg
Length of ladder = 5 m
Weight of window cleaner = 75 kg
a). Now equate the torque about the lowermost point of the ladder is given by :
[tex]$=10 \times 9.8 \times \frac{2.5}{2} + 75 \times 9.8 \times \frac{3}{5} \times 2.5 = N \times \sqrt{5^2 - 2.5^2}$[/tex]
Here, N = normal force that the glass exerts on the ladder
Therefore, [tex]$N = 282.9 \ N$[/tex]
= 280 N (in 2 significant figures)
b). Equate the forces along horizontal direction,
The horizontal component of the friction, [tex]$F_x = N = 282.9 \ N $[/tex]
The vertical component of the friction, [tex]$F_y = (10+75) \times 9.8$[/tex]
= 833 N
Therefore, the net frictional force, [tex]$F = \sqrt{F_x^2+F_y^2}$[/tex]
[tex]$F = \sqrt{(282.9)^2+(833)^2}$[/tex]
= 879.7 N
= 880 N (in 2 significant figures)
c). The angle the forces makes [tex]$= \tan \frac{833}{282.9} $[/tex]
[tex]$= 71.2 ^\circ $[/tex]
Therefore in 2 significant figures = [tex]$71 ^\circ$[/tex]
